2014 AMC 10B Problem 21

1. Draw the heights. Drop straight down from $D$ and from $C$ to the long side $AB$; call the feet $D'$ and $C'$. The angles at $A$ and $B$ are acute, so both feet land inside $AB$. The middle piece $D'C'$ equals $CD = 21$.
2. The two end pieces. Call them $x = AD'$ and $y = C'B$. Together with the middle they make up $AB$, so $x + y = 33 - 21 = 12$.
3. Pick which side is which. Say $AD = 10$ and $BC = 14$. (Swapping them just gives the mirror-image trapezoid, which has the same diagonals.)
4. Two right triangles share the height $h$. In triangle $AD'D$: $x^2 + h^2 = 10^2 = 100$. In triangle $BC'C$: $y^2 + h^2 = 14^2 = 196$.
5. Subtract to get rid of $h$: $y^2 - x^2 = 96$. Since $y^2 - x^2 = (y - x)(y + x)$ and $y + x = 12$, we get $12(y - x) = 96$, so $y - x = 8$.
6. Now $y + x = 12$ and $y - x = 8$ give $y = 10$ and $x = 2$. Then $h^2 = 100 - 2^2 = 96$.
7. The diagonals. Each diagonal is the long side of a right triangle with height $h$.
- Diagonal $AC$: from $A$ to the foot $C'$ is $33 - 10 = 23$. So $AC^2 = 23^2 + 96 = 529 + 96 = 625$, and $AC = 25$.
- Diagonal $BD$: from $B$ to the foot $D'$ is $33 - 2 = 31$. So $BD^2 = 31^2 + 96 = 961 + 96 = 1057$, which is bigger.
8. The shorter diagonal is $25$.

The answer is $\boxed{\textbf{(B)}\ 25}$.

Any trapezoid with known sides breaks into a rectangle and two right triangles that share the same height. Writing the Pythagorean theorem for both and subtracting removes the height and leaves a difference of squares, which is why the pieces come out as whole numbers. Once you know $h^2$, every other length in the picture is one more use of the Pythagorean theorem.

Slide side $DA$ to the right by $21$ so that $D$ lands on $C$. The trapezoid turns into a triangle with sides $10$, $14$ and $33 - 21 = 12$, with the same height. Heron's formula gives its area: the half-perimeter is $18$, so the area is $\sqrt{18 \cdot 8 \cdot 4 \cdot 6} = \sqrt{3456} = 24\sqrt6$. With base $12$, the height is $\frac{2 \cdot 24\sqrt6}{12} = 4\sqrt6$, so $h^2 = 96$ as before.

Drop perpendiculars from $D$ and $C$ to $\overline{AB}$, meeting it at $D'$ and $C'$. Because the angles at $A$ and $B$ are acute, both feet lie inside $\overline{AB}$, and $D'C' = CD = 21$. Let $x = AD'$ and $y = C'B$; then $x + y = 33 - 21 = 12$.

Take $AD = 10$ and $BC = 14$ (the other assignment is the mirror image and has the same diagonals). With height $h$,
$$x^2 + h^2 = 100, \qquad y^2 + h^2 = 196.$$
Subtracting, $y^2 - x^2 = 96$, so $(y - x)(y + x) = 96$ and $y - x = 8$. Hence $y = 10$, $x = 2$, and $h^2 = 100 - 4 = 96$.

Now compute the diagonals from right triangles with height $h$:

• $AC$: horizontal run $AC' = 33 - 10 = 23$, so $AC^2 = 23^2 + 96 = 529 + 96 = 625$ and $AC = 25$.
• $BD$: horizontal run $BD' = 33 - 2 = 31$, so $BD^2 = 961 + 96 = 1057$, which is larger.

The shorter diagonal is $\boxed{\textbf{(B)}\ 25}$.